Question

if x = cube root of 28 , y = cube root of 27 find x+y-1/(x²+xy+y²)
(the 5th one)​

Answer 1

$\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=6}$

Step-by-step explanation:

• Given

        $\mathbf{x=\sqrt[3]{28}}$

        So $\mathbf{x^{3}=28}$         ...1)

         $\mathbf{y=\sqrt[3]{27}=3}$

        So $\mathbf{y^{3}=27}$         ...2)

• First consider

       $\mathbf{\frac{1}{x^{2}+xy+y^{2}}}$              ...3)

      Multiply by (x-y) in numerator and denominator, we get

      $\mathbf{\frac{x-y}{(x-y)(x^{2}+xy+y^{2})}}$

      Denominator of above term can be written as

      $\mathbf{\frac{x-y}{x^{3}-y^{3}}}$          ...4)

• From equation 1), equation 2) and equation 4)

       $\mathbf{\frac{x-y}{28-27}=x-y}$    ...5)

       Means from equation 3) and equation 5)

       $\mathbf{\frac{1}{x^{2}+xy+y^{2}}=x-y}$      ...6)

• Now come to question

       $\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}}$      ...7)

       From equation 6) , equation 7) can be written as

       $\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=x+y-(x-y)}$

       $\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=x+y-x+y}$

       $\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=2y}$         (where y=3)

      $\mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=6}$

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given

        \mathbf{x=\sqrt[3]{28}}x=328

        So \mathbf{x^{3}=28}x3=28         ...1)

         \mathbf{y=\sqrt[3]{27}=3}y=327=3

        So \mathbf{y^{3}=27}y3=27         ...2)

First consider

       \mathbf{\frac{1}{x^{2}+xy+y^{2}}}x2+xy+y21              ...3)

      Multiply by (x-y) in numerator and denominator, we get

      \mathbf{\frac{x-y}{(x-y)(x^{2}+xy+y^{2})}}(x−y)(x2+xy+y2)x−y

      Denominator of above term can be written as

      \mathbf{\frac{x-y}{x^{3}-y^{3}}}x3−y3x−y          ...4)

From equation 1), equation 2) and equation 4)

       \mathbf{\frac{x-y}{28-27}=x-y}28−27x−y=x−y    ...5)

       Means from equation 3) and equation 5)

       \mathbf{\frac{1}{x^{2}+xy+y^{2}}=x-y}x2+xy+y21=x−y      ...6)

Now come to question

       \mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}}x+y−x2+xy+y21      ...7)

       From equation 6) , equation 7) can be written as

       \mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=x+y-(x-y)}x+y−x2+xy+y21=x+y−(x−y)

       \mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=x+y-x+y}x+y−x2+xy+y21=x+y−x+y

       \mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=2y}x+y−x2+xy+y21=2y         (where y=3)

      \mathbf{x+y-\frac{1}{x^{2}+xy+y^{2}}=6}x+y−x2+xy+y21=6